Question

Sunlight is incident on a surface of a glass plate at a glancing angle of 40°. The reflected ray is found to be completely plane polarized. Determine the refractive index of the glass.

Answer 1

Hint: One needs to use the concept of Brewster’s angle/angle of polarization to solve this problem. The sunlight incident on the surface of a glass plate is an unpolarized light. And when an unpolarized light incidents at Brewster’s angle, the reflected light comes out to be totally polarized, which is the same situation as in the given problem.

Complete Step-by-Step Solution:
The magnitude of the Brewster’s angle depends on the refractive index of both the optical media through which light is propagating and is given as:
\[{{\theta }_{B}}={{\tan }^{-1}}\dfrac{{{n}_{2}}}{{{n}_{1}}}\]
Where $θ_B$ is Brewster’s angle, $n_1$ is the refractive index of the medium of the incoming beam and $n_2$ the refractive index of the other medium. The sunlight is propagating from air media to glass and the refractive index of air is 1, i.e. $n_1$ = 1.

According to the question, the glancing angle is 40°, i.e. θB = 40°. We need to find \[n_2\]. Putting both the values in the above equation:
\[{{40}^{\circ }}={{\tan }^{-1}}\dfrac{{{n}_{2}}}{1}\]
\[n_2 = 0.839. \]

Hence, we can conclude that the refractive index of the glass is 0.839.

Note:
The angle between the reflected and refracted rays is 90°, when the incoming light strikes the interface at Brewster’s angle. The refracted rays in this case are partly polarized. The concept behind this condition is related to the fact that the electric field vector of the travelling light is perpendicular to the direction of propagation of the wave. When the incoming light strikes the interface, the atom at the interface absorbs that light and the electrons in the atom start oscillating in the direction of the electric field which is perpendicular to the direction of propagation of the refracted wave. The light is re-emitted by the atoms in the form of both reflected and refracted waves. When the light comes at Brewster’s angle the reflected wave has no electric field vectors parallel to the refracted wave because no electrons oscillates along that direction. And as the electric field vector won’t be parallel to the vector of propagation of reflected ray, that’s why only the perpendicular direction to the plane is possible. Hence, the reflected wave is linearly polarized.